Asked by Autumn
Use the information below to answer this question
C(s) + O2(g) -> CO2(g) ΔH = -394 kJ/mol
H2(g) + 1/2 O2(g) -> H2O(l) ΔH = -286 kJ/mol
4C(s) + 5H2(g) -> C4H10 (g) ΔH = -126 kJ/mol
The standard enthalpy of combustion of butane, in kJ/mol, is:
A -2880
B -2590
C -806
D -554
Could someone show me how to do this question? Thank you!
C(s) + O2(g) -> CO2(g) ΔH = -394 kJ/mol
H2(g) + 1/2 O2(g) -> H2O(l) ΔH = -286 kJ/mol
4C(s) + 5H2(g) -> C4H10 (g) ΔH = -126 kJ/mol
The standard enthalpy of combustion of butane, in kJ/mol, is:
A -2880
B -2590
C -806
D -554
Could someone show me how to do this question? Thank you!
Answers
Answered by
DrBob222
eqn 3 reversed + 4*eqn 1 + 5*eqn 2
Post your work if you get stuck.
Post your work if you get stuck.
Answered by
Autumn
So the answer is A, right? Thanks for the help!
Answered by
DrBob222
I didn't do the math.
Answered by
DrBob222
Yes, A is correct.
Answered by
Nintendon't
Hey! That's pretty goooood!
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